Track / Stage 5 / 25 of 400 #625 of 1964

Problem 625

AIME late

Algebra Difficulty 5.0 Find the answer HMMT_11

Let $g_{1}(x)=\frac{1}{3}\left(1+x+x^{2}+\cdots\right)$ for all values of $x$ for which the right hand side converges. Let $g_{n}(x)=g_{1}\left(g_{n-1}(x)\right)$ for all integers $n \geq 2$ . What is the largest integer $r$ such that $g_{r}(x)$ is defined for some real number $x$ ?

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

Notice that the series is geometric with ratio $x$ , so it converges if $-1<x<1$ . Also notice that where $g_{1}(x)$ is defined, it is equal to $\frac{1}{3(1-x)}$ . The image of $g_{1}(x)$ is then the interval $\left(\frac{1}{6}, \infty\right)$ . The image of $g_{2}(x)$ is simply the values of $g_{1}(x)$ for $x$ in $\left(\frac{1}{6}, 1\right)$ , which is the interval $\left(\frac{2}{5}, \infty\right)$ . Similarly, the image of $g_{3}(x)$ is $\left(\frac{5}{9}, \infty\right)$ , the image of $g_{4}(x)$ is $\left(\frac{3}{4}, \infty\right)$ , and the image of $g_{5}(x)$ is $\left(\frac{4}{3}, \infty\right)$ . As this does not intersect the interval $(-1,1), g_{6}(x)$ is not defined for any $x$ , so the answer is 5 .

Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.